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puteri [66]
3 years ago
15

Which arrangement shows 26/4 , 6.45, 6 2/5 , and 50/8 in order from least to greatest?

Mathematics
2 answers:
Lyrx [107]3 years ago
5 0
B is the correct answer.
kotegsom [21]3 years ago
4 0
It is B. Hope that helps...
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Find the volume of a cylinder with a diameter if 10 cm and a height of 10 cm?​
RideAnS [48]

Step-by-step explanation:

volume of cylinder=π(d/2)²h=π/4×10²×10=250πcm³

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1. What number is 25% of 20?
algol13

Answer:5

Step-by-step explanation:

20 divided by 4 equals 5

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Convert the exponential function to the form indicated. (Round all coefficients to four significant digits.) f(x) = 2.5 e^(-0.7
Jobisdone [24]
f(x)=2.5e^{-0.7x}=2.5\left(e^{-0.7}\right)^x\\\\A=2.5\\\\b=e^{-0.7}
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4 years ago
A smoothie recipe calls for 13 cup yogurt for each smoothie. How many cups of yogurt are needed to make 7 smoothies?
Sauron [17]
You could also say.
13 x
--- = ---
1 7
Multiply across.
1x=13×7
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Divide both sides by one.
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Or you could do it the easy way.
whatever makes you happy. :)
3 0
3 years ago
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Find the volume of the solid.
dmitriy555 [2]

In Cartesian coordinates, the region (call it R) is the set

R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}

In the plane z=2, we have

2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2

which is a circle with radius \sqrt2. Then we can better describe the solid by

R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}

so that the volume is

\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx

While doable, it's easier to compute the volume in cylindrical coordinates.

\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}

Then we can describe R in cylindrical coordinates by

R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}

so that the volume is

\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}

3 0
1 year ago
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